jmc

There is a point of intersection for each $x$ such that $f(x^2)=f(x^4)$. Since $f$ is invertible, this equation is satisfied only if $x^2=x^4$, so we simply count solutions to that equation. We can rearrange the equation $x^2=x^4$ as follows: $$\begin{array}{rl}0& =x^4-x^2\\ 0& =x^2(x^2-1)\\ 0& =x^2(x+1)(x-1)\end{array}$$The last factorization shows that the solutions are $x=-1,0,1$. Therefore, the graphs of $y=f(x^2)$ and $y=f(x^4)$ must intersect at exactly $\boxed{3}$ points.